282 Part V From the Data at Hand to the World at Large · 282 Part V From the Data at Hand to the World at Large Chapter 18 – Sampling Distribution Models 1. Send money. All of

282 Part V From the Data at Hand to the World at Large

Chapter 18 – Sampling Distribution Models

1. Send money.

All of the histograms are centered around p = 0 05. . As n gets larger, the shape of thehistograms get more unimodal and symmetric, approaching a Normal model, while thevariability in the sample proportions decreases.

2. Character recognition.

All of the histograms are centered around p = 0 85. . As n gets larger, the shapes of thehistograms get more unimodal and symmetric, approaching a Normal model, while thevariability in the sample proportions decreases.

3. Send money, again.

a)

nObserved

meanTheoretical

meanObserved

st. dev.Theoretical standard deviation

20 0.0497 0.05 0.0479 ( . )( . ) / .0 05 0 95 20 0 0487≈

50 0.0516 0.05 0.0309 ( . )( . ) / .0 05 0 95 50 0 0308≈

100 0.0497 0.05 0.0215 ( . )( . ) / .0 05 0 95 100 0 0218≈

200 0.0501 0.05 0.0152 ( . )( . ) / .0 05 0 95 200 0 0154≈

b) All of the values seem very close to what we would expect from theory.

c) The histogram for n = 200 looks quite unimodal and symmetric. We should be able to usethe Normal model.

d) The success/failure condition requires np and nq to both be at least 10, which is notsatisfied until n = 200 for p = 0.05. The theory supports the choice in part c.

4. Character recognition, again.

a)

nObserved

meanTheoretical

meanObserved


20 0.8481 0.85 0.0803 ( . )( . ) / .0 85 0 15 20 0 0799≈

50 0.8507 0.85 0.0509 ( . )( . ) / .0 85 0 15 50 0 0505≈

75 0.8481 0.85 0.0406 ( . )( . ) / .0 85 0 15 75 0 0412≈

100 0.8488 0.85 0.0354 ( . )( . ) / .0 85 0 15 100 0 0357≈

b) All of the values seem very close to what we would expect from theory.

c) Certainly, the histogram for n = 100 is unimodal and symmetric, but the histogram for n =75 looks nearly Normal, too. We should be able to use the Normal model for either.

Copyright 2010 Pearson Education, Inc.

Chapter 18 Sampling Distribution Models 283

d) The success/failure condition requires np and nq to both be at least 10, which is satisfied forboth n = 75 and n = 100 when p = 0.85. The theory supports the choice in part c.

5. Coin tosses.

a) The histogram of these proportions is expected to be symmetric, but not because of theCentral Limit Theorem. The sample of 16 coin flips is not large. The distribution of theseproportions is expected to be symmetric because the probability that the coin lands headsis the same as the probability that the coin lands tails.

b) The histogram is expected to have its center at 0.5, the probability that the coin lands heads.

c) The standard deviation of data displayed in this histogram should be approximately equal

to the standard deviation of the sampling distribution model, pq

n= =

( . )( . ).

0 5 0 516

0 125.

d) The expected number of heads, np = 16(0.5) = 8, which is less than 10. The Success/Failurecondition is not met. The Normal model is not appropriate in this case.

6. M&M’s.

a) The histogram of the proportions of green candies in the bags would probably be skewedslightly to the right, for the simple reason that the proportion of green M&M’s could neverfall below 0 on the left, but has the potential to be higher on the right.

b) The Normal model cannot be used to approximate the histogram, since the expectednumber of green M&M’s is np = 50(0.10) = 5, which is less than 10. The Success/Failurecondition is not met.

c) The histogram should be centered around the expected proportion of green M&M’s, atabout 0.10.

d) The proportion should have standard deviation pq

n= ≈

( . )( . ). .

0 1 0 950

0 042

7. More coins.

a) µ p̂ = =p 0.5 and σ ( ˆ)p = = =pq

n

( . )( . ).

0 5 0 525

0 1

About 68% of the sample proportions are expectedto be between 0.4 and 0.6, about 95% are expectedto be between 0.3 and 0.7, and about 99.7% areexpected to be between 0.2 and 0.8.

b) First of all, coin flips are independent of one another. There is no need to check the 10%Condition. Second, np = nq = 12.5, so both are greater than 10. The Success/Failurecondition is met, so the sampling distribution model is N(0.5, 0.1).



c) µ p̂ = =p 0.5 and σ ( ˆ)p = = =pq

n

( . )( . ).

0 5 0 564

0 0625

About 68% of the sample proportions are expectedto be between 0.4375 and 0.5625, about 95% areexpected to be between 0.375 and 0.625, and about99.7% are expected to be between 0.3125 and0.6875.

Coin flips are independent of one another, andnp = nq = 32, so both are greater than 10.The Success/Failure condition is met, so thesampling distribution model is N(0.5, 0.0625).

d) As the number of tosses increases, the sampling distribution model will still be Normal andcentered at 0.5, but the standard deviation will decrease. The sampling distribution modelwill be less spread out.

8. Bigger bag.

a) Randomization condition: The 200 M&M’s in thebag can be considered representative of allM&M’s, and they are thoroughly mixed.

10% condition: 200 is certainly less than 10% of allM&M’s.

Success/Failure condition: np = 20 and nq = 180are both greater than 10.

b) The sampling distribution model is Normal, with:

µ p̂ = =p 0.1 and σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 1 0 9200

0 021

About 68% of the sample proportions are expectedto be between 0.079 and 0.121, about 95% areexpected to be between 0.058 and 0.142, and about 99.7% are expected to be between 0.037and 0.163.

c) If the bags contained more candies, the sampling distribution model would still be Normaland centered at 0.1, but the standard deviation would decrease. The sampling distributionmodel would be less spread out.

9. Just (un)lucky.

For 200 flips, the sampling distribution model is Normal with µ p̂ = =p 0.5 and

σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 5 0 5200

0 0354 . Her sample proportion of ˆ .p = 0 42 is about 2.26

standard deviations below the expected proportion, which is unusual, but notextraordinary. According to the Normal model, we expect sample proportions this low orlower about 1.2% of the time.



10. Too many green ones?

For 500 candies, the sampling distribution model is Normal with µ p̂ = =p 0.1 and

σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 1 0 9500

0 01342. The sample proportion of ˆ .p = 0 12 is about 1.49

standard deviations above the expected proportion, which is not at all unusual. Accordingto the Normal model, we expect sample proportions this high or higher about 6.8% of thetime.

11. Speeding.

a) µ p̂ = =p 0.70

σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 7 0 380

0 051.

About 68% of the sample proportions are expectedto be between 0.649 and 0.751, about 95% areexpected to be between 0.598 and 0.802, and about99.7% are expected to be between 0.547 and 0.853.

b) Randomization condition: The sample may not berepresentative. If the flow of traffic is very fast, thespeed of the other cars around may have someeffect on the speed of each driver. Likewise, iftraffic is slow, the police may find a smallerproportion of speeders than they expect.

10% condition: 80 cars represent less than 10% of all cars

Success/Failure condition: np = 56 and nq = 24 are both greater than 10.

The Normal model may not be appropriate. Use caution. (And don’t speed!)

12. Smoking.

Randomization condition: 50 people are selectedat random

10% condition: 50 is less than 10% of all people.

Success/Failure condition: np = 13.2 and nq = 36.8are both greater than 10.

The sampling distribution model is Normal, with:µ p̂ = =p 0.264

σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 264 0 73650

0 062

There is an approximate chance of 68% that between 20.2% and 32.6% of 50 people aresmokers, an approximate chance of 95% that between 14.0% and 38.8% are smokers, and anapproximate chance of 99.7% that between 7.8% and 45.0%are smokers.



13. Vision.

a) Randomization condition: Assume that the 170children are a representative sample of allchildren.

10% condition: A sample of this size is less than10% of all children.

Success/Failure condition: np = 20.4 and nq = 149.6are both greater than 10.

Therefore, the sampling distribution model for theproportion of 170 children who are nearsighted isN(0.12, 0.025).

b) The Normal model is to the right.

c) They might expect that the proportion of nearsighted students to be within 2 standarddeviations of the mean. According to the Normal model, this means they might expectbetween 7% and 17% of the students to be nearsighted, or between about 12 and 29students.

14. Mortgages.

a) Randomization condition: Assume that the 1713mortgages are a representative sample of allmortgages.

10% condition: A sample of this size is less than10% of all mortgages.

Success/Failure condition: np = 27.7 and nq =1703.3 are both greater than 10.

Therefore, the sampling distribution model for theproportion of foreclosures on 1713 mortgages isN(0.016, 0.003).

b) The Normal model is to the right.

c) They might expect that the proportion of mortgage foreclosures to be within 2 standarddeviations of the mean. According to the Normal model, this means they might expectbetween 1% and 2.2% of the mortgages to undergo foreclosure, or between about 17 and 38foreclosures.

15. Loans.

a) µ p̂ = =p 7%

σ ( ˆ)p = = ≈pq

n

( . )( . ). %

0 07 0 93200

1 8



b) Randomization condition: Assume that the 200 people are a representative sample of allloan recipients.

10% condition: A sample of this size is less than 10% of all loan recipients.


Therefore, the sampling distribution model for the proportion of 200 loan recipients whowill not make payments on time is N(0.07, 0.018).

c) According to the Normal model,the probability that over 10% ofthese clients will not make timelypayments is approximately 0.048.

16. Contacts.

a) Randomization condition: 100 students are selected at random.

10% condition: 100 is less than 10% of all of the students at the university, provided theuniversity has more than 1000 students.


Therefore, the sampling distribution model for p̂ is Normal, with:

µ p̂ p= =0.30

σ ( ˆ)( . )( . )

.ppq

n= = ≈

0 30 0 70100

0 046

b) According to the Normalmodel, the probability thatmore than one-third of thestudents in this sample wearcontacts is approximately 0.234.

17. Back to school?

Randomization condition: We are considering random samples of 400 students who tookthe ACT.

10% Condition: 400 students is less than 10% of all college students.


zp

pq

n

z

z

p=−

=−

≈

ˆ

. .

( . )( . )

.

ˆµ

0 10 0 07

0 07 0 93

200

1 663

zp p

pq

n

z

z

=−

=−

≈

ˆ ˆ

.

( . )( . )

.

µ

13 0 30

0 30 0 70

100

0 727




µ p̂ p= =0.74

σ ( ˆ)( . )( . )

.ppq

n= = ≈

0 74 0 26400

0 022

According to the sampling distribution model,about 68% of the colleges are expected to haveretention rates between 0.718 and 0.762, about95% of the colleges are expected to have retentionrates between 0.696 and 0.784, and about 99.7% ofthe colleges are expected to have retention ratesbetween 0.674 and 0.806. However, theconditions for the use of this model may not bemet. We should be cautious about making anyconclusions based on this model.

18. Binge drinking.

Randomization condition: The students were selected randomly.

10% condition: 200 college students are less than 10% of all college students.



µ p̂ p= =0.44

σ ( ˆ)( . )( . )

.ppq

n= = ≈

0 44 0 56200

0 035

According to the sampling distribution model,about 68% of samples of 200 students areexpected to have binge drinking proportionsbetween 0.405 and 0.475, about 95% between0.370 and 0.510, and about 99.7% between 0.335and 0.545.

19. Back to school, again.

Provided that the students at this college are typical, the sampling distribution model forthe retention rate, p̂ , is Normal with µ p̂ p= =0.74 and standard deviation

σ ( ˆ)( . )( . )

.ppq

n= = ≈

0 74 0 26603

0 018

This college has a right to brag about their retention rate. 522/603 = 86.6% is over 7standard deviations above the expected rate of 74%.



20. Binge sample.

Since the sample is random and the Success/Failure condition is met, the samplingdistribution model for the binge drinking rate, p̂ , is Normal with µ p̂ p= =0.44 and

standard deviation σ ( ˆ)( . )( . )

.ppq

n= = ≈

0 44 0 56244

0 032

The binge drinking rate at this college is lower than the national result, but not extremelylow. 96/244 = 39.3% is only about 1.5 standard deviations below the national rate of 44%.

21. Polling.

Randomization condition: We must assume that the 400 voters were polled randomly.

10% condition: 400 voters polled represent less than 10% of potential voters.



µ p̂ p= =0.52

σ ( ˆ)( . )( . )

.ppq

n= = ≈

0 52 0 48400

0 025

According to the Normal model,the probability that thenewspaper’s sample will leadthem to predict defeat (that is,predict budget support below50%) is approximately 0.212.

22. Seeds.

Randomization condition: We must assume that the 160 seeds in a pack are a randomsample. Since seeds in each pack may not be a random sample, proceed with caution.

10% condition: The 160 seeds represent less than 10% of all seeds.

Success/Failure condition: np = 147.2 and nq = 12.8 are both greater than 10.


µ p̂ = =p 0.92

σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 92 0 08160

0 0215

According to the Normal model,the probability that more than95% of the seeds will germinateis approximately 0.081.

zp p

pq

n

z

z

=−

=−

≈ −

ˆ ˆ

. .

( . )( . )

.

µ

0 50 0 52

0 52 0 48

400

0 801

zp p

pq

n

z

z

=−

=−

≈

ˆ ˆ

. .

( . )( . )

.

µ

0 95 0 92

0 92 0 08

160

1 399



23. Apples.

Randomization condition: A random sample of 150 apples is taken from each truck.

10% condition: 150 is less than 10% of all apples.

Success/Failure Condition: np = 12 and nq = 138 are both greater than 10.


µ p̂ = =p 0.08

σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 08 0 92150

0 0222

According to the Normal model,the probability that less than 5%of the apples in the sample areunsatisfactory is approximately0.088.

24. Genetic Defect.

Randomization condition: We will assume that the 732 newborns are representative of allnewborns.

10% condition: These 732 newborns certainly represent less than 10% of all newborns.

Success/Failure condition: np = 29.28 and nq = 702.72 are both greater than 10.


µ p̂ = =p 0.04

σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 04 0 96732

0 0072

In order to get the 20 newborns for the study, the researchers hope to find at least

ˆ .p = ≈20732

0 0273 as the proportion of newborns in the sample with

juvenile diabetes.

According to the Normal model,the probability that theresearchers find at least 20newborns with juvenile diabetesis approximately 0.960.

zp p

pq

n

z

z

=−

=−

≈ −

ˆ ˆ

. .

( . )( . )

.

µ

0 05 0 08

0 08 0 92

150

1 354

zp p

pq

n

z

z

=−

=−

≈ −

ˆ ˆ

.

( . )( . )

.

µ

20732 0 04

0 04 0 96

732

1 750



25. Nonsmokers.

Randomization condition: We will assume that the 120 customers (to fill the restaurant tocapacity) are representative of all customers.

10% condition: 120 customers represent less than 10% of all potential customers.



µ p̂ = =p 0.60

σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 60 0 40120

0 0447

Answers may vary. We will use 3 standard deviations above the expected proportion ofcustomers who demand nonsmoking seats to be “very sure”.

µ ˆ . ( . ) .p

pq

n+

≈ + ≈3 0 60 3 0 0447 0 734

Since 120(0.734) = 88.08, the restaurant needs at least 89 seats in the nonsmoking section.

26. Meals.

Randomization condition: We will assume that the 180 customers are representative of allcustomers.

10% condition: 180 customers represent less than 10% of all potential customers.



µ p̂ = =p 0.20

σ ( ˆ)p = = ≈pq

n

( . )( . ).

0 20 0 80180

0 0298

Answers may vary. We will use 2 standard deviations above the expected proportion ofcustomers who order the steak special to be “pretty sure”.

µ ˆ . ( . ) .p

pq

n+

≈ + ≈2 0 20 2 0 0298 0 2596

Since 180(0.2596) = 46.728, the chef needs at least 47 steaks on hand.

27. Sampling.

a) The sampling distribution model for the sample mean is Nn

µ σ, .

b) If we choose a larger sample, the mean of the sampling distribution model will remain thesame, but the standard deviation will be smaller.



28. Sampling, part II.

a) The sampling distribution model for the sample mean will be skewed to the left as well,

centered at µ , with standard deviation σn

.

b) When the sample size is increased, the sampling distribution model becomes more Normal

in shape, centered at µ , with standard deviation σn

.

c) As we make the sample larger the distribution of data in the sample should more closelyresemble the shape, center, and spread of the population.

29. Waist size.

a) The distribution of waist size of 250 men in Utah is unimodal and slightly skewed to theright. A typical waist size is approximately 36 inches, and the standard deviation in waistsizes is approximately 4 inches.

b) All of the histograms show distributions of sample means centered near 36 inches. As ngets larger the histograms approach the Normal model in shape, and the variability in thesample means decreases. The histograms are fairly Normal by the time the sample reachessize 5.

30. CEO compensation.

a) The distribution of total compensation for the CEOs for the 800 largest U.S. companies isunimodal, but strongly skewed to the right with several large outliers.

b) All of the histograms are centered near $10,000,000. As n gets larger, the variability insample means decreases, and histograms approach the Normal shape. However, they arestill visibly skewed to the right, with the possible exception of the histogram for n = 200.

c) This rule of thumb doesn’t seem to be true for highly skewed distributions.

31. Waist size revisited.

a)

nObserved

meanTheoretical

meanObserved


2 36.314 36.33 2.855 4 019 2 2 842. / .≈5 36.314 36.33 1.805 4 019 5 1 797. / .≈10 36.341 36.33 1.276 4 019 10 1 271. / .≈20 36.339 36.33 0.895 4 019 20 0 897. / .≈

b) The observed values are all very close to the theoretical values.

c) For samples as small as 5, the sampling distribution of sample means is unimodal andsymmetric. The Normal model would be appropriate.

d) The distribution of the original data is nearly unimodal and symmetric, so it doesn’t take avery large sample size for the distribution of sample means to be approximately Normal.



32. CEOs revisited.

a)

nObserved

meanTheoretical

meanObserved


30 10251.73 10307.31 3359.64 17964 62 30 3279 88. / .≈50 10343.93 10307.31 2483.84 17964 62 50 2540 58. / .≈100 10329.94 10307.31 1779.18 17964 62 100 1796 46. / .≈200 10340.37 10307.31 1260.79 17964 62 200 1270 29. / .≈

b) The observed values are all very close to the theoretical values.

c) All the sampling distributions are still quite skewed, with the possible exception of thesampling distribution for n = 200, which is still somewhat skewed. The Normal modelwould not be appropriate.

d) The distribution of the original data is strongly skewed, so it will take a very large samplesize before the distribution of sample means is approximately Normal.

33. GPAs.

Randomization condition: Assume that thestudents are randomly assigned to seminars.

Independence assumption: It is reasonable tothink that GPAs for randomly selected studentsare mutually independent.

10% condition: The 25 students in the seminarcertainly represent less than 10% of the populationof students.

Large Enough Sample condition: The distributionof GPAs is roughly unimodal and symmetric, sothe sample of 25 students is large enough.

The mean GPA for the freshmen was µ = 3 4. , with standard deviation σ = 0 35. . Since theconditions are met, the Central Limit Theorem tells us that we can model the samplingdistribution of the mean GPA with a Normal model, with µy = 3 4. and standard deviation

σ ( ).

. .y = ≈0 3525

0 07

The sampling distribution model for the sample mean GPA is approximately N( . , . )3 4 0 07 .



34. Home values.

Randomization condition: Homes were selected at random.

Independence assumption: It is reasonable to think that assessments for randomlyselected homes are mutually independent.

10% condition: The 100 homes in the sample certainly represent less than 10% of thepopulation of all homes in the city. A small city will likely have more than 1,000 homes.

Large Enough Sample condition: A sample of 100 homes is large enough.

The mean home value was µ = $ ,140 000, withstandard deviation σ = $ ,60 000. Since theconditions are met, the Central Limit Theorem tellsus that we can model the sampling distribution ofthe mean home value with a Normal model, withµy = $ ,140 000 and standard deviation

σ ( ),

$ .y = =60 000100

6000

The sampling distribution model for the samplemean home values is approximatelyN( , )140000 6000 .

35. Lucky spot?

a) Smaller outlets have more variability than the larger outlets, just as the Central LimitTheorem predicts.

b) If the lottery is truly random, all outlets are equally likely to sell winning tickets.

36. Safe cities.

The standard deviation of the sampling model for the mean is

σn

. So, cities in which the

average is based on a smaller number of drivers will have greater variation in theiraverages and will be more likely to be both safest and least safe.

37. Pregnancy.

a)z

y

z

z

=−

=−

=

µσ

270 26616

0 25.

zy

z

z

=−

=−

=

µσ

280 26616

0 875.

According to theNormal model,approximately 21.1%of all pregnancies areexpected to lastbetween 270 and 280days.



b)

c) Randomization condition: Assume that the 60 women the doctor is treating can beconsidered a representative sample of all pregnant women.

Independence assumption: It is reasonable to think that the durations of the patients’pregnancies are mutually independent.

10% condition: The 60 women that the doctor is treating certainly represent less than 10%of the population of all women.

Large Enough Sample condition: The sample of 60 women is large enough. In this case,any sample would be large enough, since the distribution of pregnancies is Normal.

The mean duration of the pregnancies was µ = 266 days, with standard deviationσ = 16 days. Since the distribution of pregnancy durations is Normal, we can model thesampling distribution of the mean pregnancy duration with a Normal model, with

µy = 266 days and standard deviation σ ( ) .y = ≈1660

2 07 days .

d) According to the Normal model, the probabilitythat the mean pregnancy duration is less than 260days is 0.002.

38. Rainfall.

a) According to the Normal model, Ithaca is expectedto get more than 40 inches of rain in approximately13.7% of years.

zy

y

y

=−

=−

≈

µσ

0 674266

16276 8

.

. days

According to the Normalmodel, the longest 25% ofpregnancies are expected tolast approximately 276.8days or more.



b) According to the Normalmodel, Ithaca is expected to getless than 31.9 inches of rain indriest 20% of years.

c) Randomization condition: Assume that the 4 years in which the student was in Ithaca canbe considered a representative sample of all years.

Independence assumption: It is reasonable to think that the rainfall totals for the years aremutually independent.

10% condition: The 4 years in which the student was in Ithaca certainly represent less than10% of all years.

Large enough sample condition: The sample of 4 years is large enough. In this case, anysample would be large enough, since the distribution of annual rainfall is Normal.

The mean rainfall was µ = 35 4. inches, with standard deviation σ = 4 2. inches. Since thedistribution of yearly rainfall is Normal, we can model the sampling distribution of themean annual rainfall with a Normal model, with µy = 35 4. inches and standard deviation

σ ( ).

.y = =4 24

2 1 inches .

d) According to the Normal model, the probabilitythat those four years averaged less than 30 inchesof rain is 0.005.

39. Pregnant again.

a) The distribution of pregnancy durations may be skewed to the left since there are morepremature births than very long pregnancies. Modern practice of medicine stopspregnancies at about 2 weeks past normal due date by inducing labor or performing aCaesarean section.

b) We can no longer answer the questions posed in parts a and b. The Normal model is notappropriate for skewed distributions. The answer to part c is still valid. The Central LimitTheorem guarantees that the sampling distribution model is Normal when the sample sizeis large.

40. At work.

a) The distribution of length of time people work at a job is likely to be skewed to the right,because some people stay at the same job for much longer than the mean plus two or threestandard deviations. Additionally, the left tail cannot be long, because a person cannotwork at a job for less than 0 years.

zy

y

y

= −

− = −

≈

µσ

0 84235 4

4 231 9

..

..



b) The Central Limit Theorem guarantees that the distribution of the mean time is Normallydistributed for large sample sizes, as long as the assumptions and conditions are satisfied.The CLT doesn’t help us with the distribution of individual times.

41. Dice and dollars.

a) Let X = the number of dollars won in one play.

µ = =

+

+

=E X( $) 0

36

126

1016

2

σ

σ

2 2 2 20 236

1 226

10 216

13

13 3 61

= = −

+ −

+ −

=

= = = ≈

Var X

SD X Var X

( ) ( ) ( ) ( )

( ) ( ) $ .

b) X + X = the total winnings for two plays.

µ

σ

= + = + = + =

= + = +

= + ≈

E X X E X E X

SD X X Var X Var X

( ) ( ) ( ) $

( ) ( ) ( )

$ .

2 2 4

13 13 5 10

c) In order to win at least $100 in 40 plays, you must average at least 10040

2 50= $ . per play.

The expected value of the winnings is µ = $2, with standard deviation σ = $ .3 61. Rolling adie is random and the outcomes are mutually independent, so the Central Limit Theoremguarantees that the sampling distribution model is Normal with µx = $2 and standard

deviation σ ( )$ .

$ .x = ≈3 6140

0 571.

According to the Normal model, the probabilitythat you win at least $100 in 40 plays isapproximately 0.191.

(This is equivalent to using N(80, 22.83) to modelyour total winnings.)

42. New game.

a) Let X = the amount of money won.

b) µ = =

+

−

≈ −E X( $ .) 40

16

0536

102536

0 28

σ

σ

2 2 2 240 0 2816

0 0 28536

10 0 282536

336 034

336 034 18 33

= = − −

+ − −

+ − − −

≈

= = = ≈

Var X

SD X Var X

( ) ( ( . )) ( ( . )) ( ( . )) .

( ) ( ) . $ .

X $40 $0 – $10

P(X)16

56

16

536

= 56

56

2536

=



c) µ = + + + + = = − = −E X X X X X E X( ( ) ( . ) $ .) 5 5 0 28 1 40

σ = + + + + = = ≈SD X X X X X Var X( ) ( ( )) ( . ) $ .5 5 336 034 40 99

d) In order for the person running the game to make a profit, the average winnings of the 100people must be less than $0.

The expected value of the winnings is µ = −$ .0 28, with standard deviation σ = $ .18 33.Rolling a die is random and the outcomes are mutually independent, so the Central LimitTheorem guarantees that the sampling distribution model is Normal with µx = −$ .0 28 and

standard deviation σ ( ).

$ .x = ≈18 33100

1 833 .

According to the Normal model, the probabilitythat the person running the game makes a profit isapproximately 0.561.

43. AP Stats 2006.

a) µ = + + + +5 0 126 4 0 222 3 0 253 2 0 183 1 0 2( . ) ( . ) ( . ) ( . ) ( . 116 2 859) .≈

σ =− + − + −( . ) ( . ) ( . ) ( . ) ( .5 2 859 0 126 4 2 859 0 222 3 22 2 8859 0 253

2 2 859 0 183 1 2 859

2

2 2

) ( . )

( . ) ( . ) ( . ) (+ − + − 00 2161 324

. ).≈

The calculation for standard deviation is based on a rounded mean. Use technology tocalculate the mean and standard deviation to avoid inaccuracy.

b) The distribution of scores for 40 randomly selected students would not follow a Normalmodel. The distribution would resemble the population, mostly uniform for scores 1 – 4,with about half as many 5s.

c) Randomization condition: The scores are selected randomly.

Independence assumption: It is reasonable to think that the randomly selected scores areindependent of one another.

10% condition: The 40 scores represent less than 10% of all scores.

Large Enough Sample condition: A sample of 40 scores is large enough.

Since the conditions are satisfied, the sampling distribution model for the mean of 40randomly selected AP Stat scores is Normal, with µ µ

y= ≈ 2 859. and standard deviation

σ σ( )

..y

n= = ≈

1 324

400 2093 .



44. Museum membership.

a) µ = + + + + ≈50 0 41 100 0 37 250 0 14 500 0 07 1000 0 01 137 50( . ) ( . ) ( . ) ( . ) ( . ) $ .

σ =− + − + −

+ − + −≈

( . ) ( . ) ( . ) ( . ) ( . ) ( . )

( . ) ( . ) ( . ) ( . )$ .

50 137 50 0 41 100 137 50 0 37 250 137 50 0 14

500 137 50 0 07 1000 137 50 0 01148 56

2 2 2

2 2

The calculation for standard deviation is based on a rounded mean. Use technology tocalculate the mean and standard deviation to avoid inaccuracy.

b) The distribution of donations for 50 new members would not follow a Normal model. Thenew members would probably make donations typical of the current member populations,so the distribution would resemble the population, skewed to the right.

c) Randomization condition: The members are not selected randomly. They are simply thenew members that day. However, the donations they make are probably typical of thedonations made by current members.

Independence assumption: It is reasonable to think that the donations for the newmembers would not affect one another.

10% condition: The 50 donations represent less than 10% of all donations.

Large Enough Sample condition: The sample of 50 donations is large enough.

Since the conditions are satisfied, the sampling distribution model for the mean of 50

donations is Normal, with µ µy = ≈ $ .137 50 and standard deviation σ σ( )

.y

n= =

148 5650

.

45. AP Stats 2006, again.

Since the teacher considers his 63 students “typical”, and 63 is less than 10% of all students,the sampling distribution model for the mean AP Stat score for 63 students is Normal, with

mean µ µy

= ≈ 2 859. and standard deviation σ σ( )

..y

n= = ≈

1 324

630 167 .

zy

y

z

z

y=−

=−

=

µσ ( )

.

..

3 2 8590 167

0 844

According to the samplingdistribution model, theprobability that the class of 63students achieves an average of3 on the AP Stat exam is about20%.



46. Joining the museum.

If the new members can be considered a random sample of all museum members, and the80 new members are less than 10% of all members, then the sampling distribution modelfor the mean donation of 80 members is Normal, with µ µy = ≈ $ .137 50 and standard

deviation σ ( ).

$ .y = =148 56

8016 61.

47. Pollution.

a) Randomization condition: Assume that the 80 cars can be considered a representativesample of all cars of this type.

Independence assumption: It is reasonable to think that the CO emissions for these carsare mutually independent.

10% condition: The 80 cars in the fleet certainly represent less than 10% of all cars of thistype.

Large Enough Sample condition: A sample of 80 cars is large enough.

The mean CO level was µ = 2 9. gm/mi, with standard deviation σ = 0 4. gm/mi. Since theconditions are met, the CLT allows us to model the sampling distribution of the y with a

Normal model, with µy = 2 9. gm/mi and standard deviation σ ( ).

.y = =0 480

0 045 gm/mi.

b) According to the Normal model, theprobability that y is between 3.0 and3.1 gm/mi is approximately 0.0131.

c) According to the Normal model,there is only a 5% chance that thefleet’s mean CO level is greaterthan approximately 2.97 gm/mi.

zy

y

y

y

y=−

= −

≈

µσ ( )

..

..

1 6452 9

0 0452 97

zy

y

z

z

y=−

=−

= −

µσ ( )

..

.

100 137 5016 61

2 258

According to the samplingdistribution model, there isa 98.8% probability that theaverage donation for 80new members is at least$100.



48. Potato chips.

a) According to the Normal model, only about 4.78%of the bags sold are underweight.

b) P(none of the 3 bags are underweight)= − ≈( . ) . .1 0 0478 0 8633

c) Randomization condition: Assume that the 3bags can be considered a representative sample ofall bags.

Independence assumption: It is reasonable to think that the weights of these bags aremutually independent.

10% condition: The 3 bags certainly represent less than 10% of all bags.

Large Enough Sample condition: Since the distribution of bag weights is believed to beNormal, the sample of 3 bags is large enough.

The mean weight is µ = 10 2. ounces, with standard deviation σ = 0 12. ounces. Since theconditions are met, we can model the sampling distribution of y with a Normal model,with µy = 10 2. ounces and standard deviation

σ ( ).

.y = ≈0 123

0 069 ounces.

According to the Normal model, the probabilitythat the mean weight of the 3 bags is less than 10ounces is approximately 0.0019.

d) For 24 bags, the standard deviation of the sampling distribution model is

σ ( ).

.y = ≈0 1224

0 024 ounces. Now, an average of 10 ounces is over 8 standard deviations

below the mean of the sampling distribution model. This is extremely unlikely.

49. Tips.

a) Since the distribution of tips is skewed to the right, we can’t use the Normal model todetermine the probability that a given party will tip at least $20.

b) No. A sample of 4 parties is probably not a large enough sample for the CLT to allow us touse the Normal model to estimate the distribution of averages.

c) A sample of 10 parties may not be large enough to allow the use of a Normal model todescribe the distribution of averages. It would be risky to attempt to estimate theprobability that his next 10 parties tip an average of $15. However, since the distribution oftips has µ = $ .9 60, with standard deviation σ = $ .5 40, we still know that the mean of the

sampling distribution model is µy = $ .9 60 with standard deviation σ ( ).

$ . .y = ≈5 4010

1 71

We don’t know the exact shape of the distribution, but we can still assess the likelihood ofspecific means. A mean tip of $15 is over 3 standard deviations above the expected meantip for 10 parties. That’s not very likely to happen.



50. Groceries.

a) Since the distribution of Sunday purchases is skewed, we can’t use the Normal model todetermine the probability that a given purchase is at least $40.

b) A sample of 10 customers may not be large enough for the CLT to allow the use of aNormal model for the sampling distribution model. If the distribution of Sundaypurchases is only slightly skewed, the sample may be large enough, but if the distributionis heavily skewed, it would be very risky to attempt to determine the probability.

c) Randomization condition: Assume that the 50 Sunday purchases can be considered arepresentative sample of all purchases.

Independence assumption: It is reasonable to think that the Sunday purchases aremutually independent, unless there is a sale or other incentive to purchase more.

10% condition: The 50 purchases certainly represent less than 10% of all purchases.

Large Enough Sample condition: The sample of 50 purchases is large enough.

The mean Sunday purchase is µ = $32, with standard deviation σ = $20. Since theconditions are met, the CLT allows us to model the sampling distribution of y with aNormal model, with µy = $32 and standard

deviation σ ( ) $ .y = ≈2050

2 83.

According to the Normal model, theprobability the mean Sunday purchase of 50customers is at least $40 is about 0.0023.

51. More tips.

a) Randomization condition: Assume that the tips from 40 parties can be considered arepresentative sample of all tips.

Independence assumption: It is reasonable to think that the tips are mutuallyindependent, unless the service is particularly good or bad during this weekend.

10% condition: The tips of 40 parties certainly represent less than 10% of all tips.

Large Enough Sample condition: The sample of 40 parties is large enough.

The mean tip is µ = $ .9 60, with standard deviation σ = $ .5 40 . Since the conditions aresatisfied, the CLT allows us to model the sampling distribution of y with a Normal model,

with µy = $ .9 60 and standard deviation σ ( ).

$ . .y = ≈5 4040

0 8538

In order to earn at least $500, the waiter would

have to average 50040

12 50= $ . per party.

According to the Normal model, theprobability that the waiter earns at least $500in tips in a weekend is approximately 0.0003.



b) According to the Normal model,the waiter can expect to have amean tip of about $10.6942,which corresponds to about$427.77 for 40 parties, in the best10% of such weekends.

52. More groceries.

a) Assumptions and conditions for the use of the CLT were verified in a previous exercise.

The mean purchase is µ = $32, with standard deviation σ = $20. Since the sample is large,the CLT allows us to model the sampling distribution of y with a Normal model, with

µy = $32 and standard deviation σ ( ) $ .y = ≈20312

1 1323.

In order to have revenues of at least $10,000, the mean Sunday purchase must be at least10 000

31232 0513

,$ .= .

According to the Normal model, the probability ofhaving a mean Sunday purchase at least that high(and therefore at total revenue of at least $10,000)is 0.482.

b) According to the Normalmodel, the mean Sundaypurchase on the worst 10%of such days isapproximately $30.548928,so 312 customers areexpected to spend about$9531.27.

53. IQs.

a) According to the Normal model, the probabilitythat the IQ of a student from East State is at least125 is approximately 0.734.

zy

y

y

y

y=−

= −

≈

µσ ( )

..

.

.

1 28169 60

0 853810 6942

zy

y

y

y

y=−

− = −

≈

µσ ( )

.

.

1 28155232

20312

30 548928



b) First, we will need to generate a model for thedifference in IQ between the two schools. Since weare choosing at random, it is reasonable to believethat the students’ IQs are independent, whichallows us to calculate the standard deviation of thedifference.

µ

σ

= − = − = − =

= − = +

= + ≈

E E W E E E W

SD E W Var E Var W

( ) ( ) ( )

( ) ( ) ( )

.

130 120 10

8 10 12 8062 2

Since both distributions are Normal, the distribution of the difference is N(10, 12.806).

According to the Normal model, the probability that the IQ of a student at ESU is at least 5points higher than a student at WSU is approximately 0.652.

c) Randomization condition: Students are randomly sampled from WSU.

Independence assumption: It is reasonable to think that the IQs are mutuallyindependent.

10% condition: The 3 students certainly represent less than 10% of students.

Large Enough Sample condition: The distribution of IQs is Normal, so the distribution ofsample means of samples of any size will be Normal, so a sample of 3 students is largeenough.

The mean IQ is µw = 120, with standard deviationσ w = 10 . Since the distribution IQs is Normal, wecan model the sampling distribution of w with aNormal model, with µw = 120 with standard

deviation σ ( ) . .w = ≈103

5 7735

According to the Normal model, the probabilitythat the mean IQ of the 3 WSU students is above 125 is approximately 0.193.

d) As in part c, the sampling distribution of e , the mean IQ of 3 ESU students, can be

modeled with a Normal model, with µe = 130 with standard deviation σ ( ) . .e = ≈83

4 6188

The distribution of the difference in mean IQ is Normal, with the following parameters:

µ

σe w

e w

E e w E e E w

SD e w Var e Var w

−

−

= − = − = − =

= − = +

=

+

≈

( ) ( ) ( )

( ) ( ) ( )

.

130 120 10

103

83

7 39372 2



According to the Normal model, theprobability that the mean IQ of 3 ESU studentsis at least 5 points higher than the mean IQ of 3WSU students is approximately 0.751.

54. Milk.

a) According to the Normal model, the probabilitythat an Ayrshire averages more than 50 pounds ofmilk per day is approximately 0.309.

b) First, we will need to generate a model for the difference in milk production between thetwo cows. Since we are choosing at random, it is reasonable to believe that the cows’ milkproductions are independent, which allows us to calculate the standard deviation of thedifference.

µ

σ

= − = − = − =

= − = +

= + ≈

E A J E A E J

SD A J Var A Var J

( ) ( ) ( )

( ) ( ) ( )

.

47 43 4

6 5 7 81022 2

pounds

pounds

Since both distributions are Normal, thedistribution of the difference is N(4, 7.8102).

According to the Normal model, the probabilitythat the Ayrshire gives more milk than the Jerseyis approximately 0.696.

c) Randomization condition: Assume that thefarmer’s 20 Jerseys are a representative sample of all Jerseys.

Independence assumption: It is reasonable to think that the cows have mutuallyindependent milk production.

10% condition: The 20 cows certainly represent less than 10% of cows.

Large Enough Sample condition: Since the distribution of daily milk production isNormal, the sample means of samples of any size are Normally distributed, so the sampleof 20 cows is certainly large enough.



The mean milk production is µ j = 43 pounds , with standard deviation σ j = 5. Since thedistribution of milk production is Normal, we can model the sampling distribution of jwith a Normal model, with µ

j= 43 pounds with

standard deviation σ ( ) .j = ≈520

1 1180 .

According to the Normal model, the probabilitythat the mean milk production of the 20 Jerseys isabove 45 pounds of milk per day is approximately0.037.

d) As in part c, the sampling distribution of a , the mean milk production of 10 Ayrshires, canbe modeled with a Normal model, with µa = 47 pounds with standard deviation

σ ( ) .a = ≈610

1 8974 pounds.

The distribution of the difference in mean milk production is Normal, with the followingparameters:

µ

σ

a j

a j

E a j E a E j

SD a j Var a Var j

−

−

= − = − = − =

= − = +

=

+

≈

( ) ( ) ( )

( ) ( ) ( )

.

47 43 4

610

520

2 20232 2

pounds

pounds

According to the Normal model, the probabilitythat the mean milk production of 10 Ayrshires isat least 5 pounds higher than the mean milkproduction of 20 Jerseys is approximately 0.325.


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282 Part V From the Data at Hand to the World at Large · 282 Part V From the Data at Hand to the World at Large Chapter 18 – Sampling Distribution Models 1. Send money. All of